Γ
(
z
)
{\displaystyle \,\Gamma (z)\,}
f
(
z
+
1
)
=
z
f
(
z
)
{\displaystyle \,f(z+1)=zf(z)\,}
log
(
f
(
z
)
)
{\displaystyle \,\log(f(z))\,}
f
(
1
)
=
1
{\displaystyle \,f(1)=1\,}
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
Γ
(
x
+
1
)
=
x
Γ
(
x
)
{\displaystyle \,\Gamma (x+1)=x\Gamma (x)\,}
log
(
Γ
(
x
)
)
{\displaystyle \,\log \left(\Gamma (x)\right)\,}
Γ
(
1
)
=
1
{\displaystyle \,\Gamma (1)=1\,}
Aslında
Γ
(
x
+
1
)
=
x
Γ
(
x
)
{\displaystyle \,\Gamma (x+1)=x\Gamma (x)\,}
Γ
(
x
+
n
)
=
(
x
+
n
−
1
)
(
x
+
n
−
2
)
(
x
+
n
−
3
)
…
(
x
+
1
)
x
Γ
(
x
)
{\displaystyle \,{\begin{aligned}\Gamma (x+n)=(x+n-1)(x+n-2)(x+n-3)\ldots (x+1)x\Gamma (x)\end{aligned}}\,}
ve bu sonuçtan hareketle
Γ
(
1
)
=
1
{\displaystyle \,\Gamma (1)=1\,}
Γ
(
x
+
1
)
=
x
Γ
(
x
)
{\displaystyle \,\Gamma (x+1)=x\Gamma (x)\,}
Γ
(
n
)
=
(
n
−
1
)
!
{\displaystyle \,\Gamma (n)=(n-1)!\,}
n
∈
N
{\displaystyle \,n\in \mathbb {N} \,}
ve eğer
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
0
<
x
≤
1
{\displaystyle \,0<x\leq 1\,}
Γ
(
x
+
n
)
{\displaystyle \,\Gamma (x+n)\,}
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
x
{\displaystyle \,x\,}
(
x
1
,
f
(
x
1
)
)
{\displaystyle \,(x_{1},\;f(x_{1}))\,}
(
x
2
,
f
(
x
2
)
)
{\displaystyle \,(x_{2},\;f(x_{2}))\,}
x
1
<
x
2
{\displaystyle \,x_{1}<x_{2}\,}
M
(
x
1
,
x
2
)
{\displaystyle \,{\mathcal {M}}(x_{1},x_{2})\,}
konveks fonksiyon ile onun doğal öngörüsüden dolayı
log
(
Γ
(
x
)
)
{\displaystyle \,\log \left(\Gamma (x)\right)\,}
M
(
n
−
1
,
n
)
≤
M
(
n
,
n
+
x
)
≤
M
(
n
,
n
+
1
)
w
h
e
n
0
<
x
≤
1
log
(
Γ
(
n
−
1
)
)
−
log
(
Γ
(
n
)
)
(
n
−
1
)
−
n
≤
log
(
Γ
(
n
)
)
−
log
(
Γ
(
n
+
x
)
)
n
−
(
n
+
x
)
≤
log
(
Γ
(
n
)
)
−
log
(
Γ
(
n
+
1
)
)
n
−
(
n
+
1
)
log
(
(
n
−
2
)
!
)
−
log
(
(
n
−
1
)
!
)
−
1
≤
log
(
Γ
(
n
+
x
)
)
−
log
(
(
n
−
1
)
!
)
x
≤
log
(
n
!
)
−
log
(
(
n
−
1
)
!
)
1
−
log
(
(
n
−
2
)
!
(
n
−
1
)
!
)
≤
log
(
Γ
(
n
+
x
)
)
−
log
(
(
n
−
1
)
!
)
x
≤
log
(
n
!
(
n
−
1
)
!
)
−
log
(
1
(
n
−
1
)
)
≤
log
(
Γ
(
n
+
x
)
)
−
log
(
(
n
−
1
)
!
)
x
≤
log
(
n
)
x
⋅
log
(
n
−
1
)
+
log
(
(
n
−
1
)
!
)
≤
log
(
Γ
(
n
+
x
)
)
≤
x
⋅
log
(
n
)
+
log
(
(
n
−
1
)
!
)
log
(
(
n
−
1
)
x
(
n
−
1
)
!
)
≤
log
(
Γ
(
n
+
x
)
)
≤
log
(
n
x
(
n
−
1
)
!
)
{\displaystyle {\begin{aligned}{\mathcal {M}}(n-1,n)&\leq {\mathcal {M}}(n,n+x)\leq {\mathcal {M}}(n,n+1)\;\;\mathrm {when} \;0<x\leq 1\\{\frac {\log \left(\Gamma (n-1)\right)-\log \left(\Gamma (n)\right)}{(n-1)-n}}&\leq {\frac {\log \left(\Gamma (n)\right)-\log \left(\Gamma (n+x)\right)}{n-(n+x)}}\leq {\frac {\log \left(\Gamma (n)\right)-\log \left(\Gamma (n+1)\right)}{n-(n+1)}}\\{\frac {\log \left((n-2)!\right)-\log \left((n-1)!\right)}{-1}}&\leq {\frac {\log \left(\Gamma (n+x)\right)-\log \left((n-1)!\right)}{x}}\leq {\frac {\log \left(n!\right)-\log \left((n-1)!\right)}{1}}\\-\log \left({\frac {(n-2)!}{(n-1)!}}\right)&\leq {\frac {\log \left(\Gamma (n+x)\right)-\log \left((n-1)!\right)}{x}}\leq \log \left({\frac {n!}{(n-1)!}}\right)\\-\log \left({\frac {1}{(n-1)}}\right)&\leq {\frac {\log \left(\Gamma (n+x)\right)-\log \left((n-1)!\right)}{x}}\leq \log \left(n\right)\\x\cdot \log \left(n-1\right)+\log \left((n-1)!\right)&\leq \log \left(\Gamma (n+x)\right)\leq x\cdot \log \left(n\right)+\log \left((n-1)!\right)\\\log \left((n-1)^{x}(n-1)!\right)&\leq \log \left(\Gamma (n+x)\right)\leq \log \left(n^{x}(n-1)!\right)\end{aligned}}}
Böyle bir limit varlığı veya yakınsama gibi çeşitli şeyleri kanıtlamak için ortak bir analiz tekniğidir.
Şimdi biz bu fonksiyonu geri çağırıyoruz ve her ikisi monoton artandır .
Bu, iki ifade arasında sıkışmış olan fonksiyon son satırından bellidir
log
(
)
{\displaystyle \,\log()\,}
e
(
)
{\displaystyle \,e^{()}\,}
(
n
−
1
)
x
(
n
−
1
)
!
≤
Γ
(
n
+
x
)
≤
n
x
(
n
−
1
)
!
(
n
−
1
)
x
(
n
−
1
)
!
≤
(
x
+
n
−
1
)
(
x
+
n
−
2
)
…
(
x
+
1
)
x
Γ
(
x
)
≤
n
x
(
n
−
1
)
!
(
n
−
1
)
x
(
n
−
1
)
!
(
x
+
n
−
1
)
(
x
+
n
−
2
)
…
(
x
+
1
)
x
≤
Γ
(
x
)
≤
n
x
(
n
−
1
)
!
(
x
+
n
−
1
)
(
x
+
n
−
2
)
…
(
x
+
1
)
x
(
n
−
1
)
x
(
n
−
1
)
!
(
x
+
n
−
1
)
(
x
+
n
−
2
)
…
(
x
+
1
)
x
≤
Γ
(
x
)
≤
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
(
n
+
x
n
)
{\displaystyle \,{\begin{aligned}(n-1)^{x}(n-1)!&\leq \Gamma (n+x)\leq n^{x}(n-1)!\\(n-1)^{x}(n-1)!&\leq (x+n-1)(x+n-2)\ldots (x+1)x\Gamma (x)\leq n^{x}(n-1)!\\{\frac {(n-1)^{x}(n-1)!}{(x+n-1)(x+n-2)\ldots (x+1)x}}\leq \Gamma (x)&\leq {\frac {n^{x}(n-1)!}{(x+n-1)(x+n-2)\ldots (x+1)x}}\\{\frac {(n-1)^{x}(n-1)!}{(x+n-1)(x+n-2)\ldots (x+1)x}}&\leq \Gamma (x)\leq {\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\left({\frac {n+x}{n}}\right)\\\end{aligned}}\,}
Son satırı güçlü bir ifadedir.
Özelde, bütün
n
{\displaystyle \,n\,}
de geçerlidir.
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
n
{\displaystyle \,n\,}
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
n
{\displaystyle \,n\,}
n
+
1
{\displaystyle \,n+1\,}
n
{\displaystyle \,n\,}
(
(
n
+
1
)
−
1
)
x
(
(
n
+
1
)
−
1
)
!
(
x
+
(
n
+
1
)
−
1
)
(
x
+
(
n
+
1
)
−
2
)
…
(
x
+
1
)
x
≤
Γ
(
x
)
≤
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
(
n
+
x
n
)
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
≤
Γ
(
x
)
≤
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
(
n
+
x
n
)
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
≤
Γ
(
x
)
Γ
(
x
)
(
n
n
+
x
)
≤
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
{\displaystyle \,{\begin{aligned}{\frac {((n+1)-1)^{x}((n+1)-1)!}{(x+(n+1)-1)(x+(n+1)-2)\ldots (x+1)x}}&\leq \Gamma (x)\leq {\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\left({\frac {n+x}{n}}\right)\\{\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}&\leq \Gamma (x)\leq {\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\left({\frac {n+x}{n}}\right)\\{\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}&\leq \Gamma (x)\\\Gamma (x)\left({\frac {n}{n+x}}\right)&\leq {\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\end{aligned}}\,}
Bu son iki ifadeyi birleştirirsek
Γ
(
x
)
(
n
n
+
x
)
≤
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
≤
Γ
(
x
)
{\displaystyle \,{\begin{aligned}\Gamma (x)\left({\frac {n}{n+x}}\right)\leq {\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\leq \Gamma (x)\end{aligned}}\,}
şimdi
n
→
∞
{\displaystyle \,n\rightarrow \infty \,}
n
n
+
x
→
1
{\displaystyle \,{\frac {n}{n+x}}\rightarrow 1\,}
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
{\displaystyle \,{\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\,}
lim
n
→
∞
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
{\displaystyle \,\lim _{n\rightarrow \infty }{\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\,}
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
lim
n
→
∞
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
{\displaystyle \,\lim _{n\rightarrow \infty }{\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\,}
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
dizinin limiti benzersiz olduğu hatırdan çıkarılmamalıdır
Bu demektir ki herhangi bir
x
∈
(
0
,
1
]
{\displaystyle \,x\in (0,1]\,}
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
ispat sorusunun teorem varsayımı kalan diğer ucudur
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
x
{\displaystyle \,x\,}
lim
n
→
∞
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
{\displaystyle \,\lim _{n\rightarrow \infty }{\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\,}
M
(
n
−
1
,
n
)
≤
M
(
n
+
x
,
n
)
≤
M
(
n
+
1
,
n
)
{\displaystyle \,{\begin{aligned}{\mathcal {M}}(n-1,n)\leq {\mathcal {M}}(n+x,n)\leq {\mathcal {M}}(n+1,n)\end{aligned}}\,}
0
<
x
≤
1
{\displaystyle \,0<x\leq 1\,}
x
>
1
{\displaystyle \,x>1\,}
M
{\displaystyle \,{\mathcal {M}}\,}
M
(
n
+
1
,
n
)
<
M
(
n
+
x
,
n
)
{\displaystyle \,{\mathcal {M}}(n+1,n)<{\mathcal {M}}(n+x,n)\,}
Γ
(
x
+
1
)
=
lim
n
→
∞
x
⋅
(
n
x
n
!
(
x
+
n
)
(
x
+
n
−
1
)
…
(
x
+
1
)
x
)
n
n
+
x
+
1
Γ
(
x
)
=
(
1
x
)
Γ
(
x
+
1
)
{\displaystyle \,{\begin{aligned}\Gamma (x+1)&=\lim _{n\rightarrow \infty }x\cdot \left({\frac {n^{x}n!}{(x+n)(x+n-1)\ldots (x+1)x}}\right){\frac {n}{n+x+1}}\\\Gamma (x)&=\left({\frac {1}{x}}\right)\Gamma (x+1)\end{aligned}}\,}
dikkat edilmelidir.
ilk olarak gösterilen
x
{\displaystyle \,x\,}
Γ
(
x
)
{\displaystyle \,\Gamma (x)\,}
![{\displaystyle \,x\in (0,1]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bf918407a2b1e5250adf9e7bca1dedf9f475e2b)
